First Order Differential Equations - Making a substitution

Question

I was wondering if I could get some advice on how to tackle this question:

Consider the differential equation

$$x^2 {dy\over dx}+2xy- y^3 = 0 \quad (3)$$

Make the substitution $u=y^{-2}$ and show that the differential equation reduces to

$$ -{1 \over 2}x^2{du\over dx}+2xu-1=0 \quad (4)$$

Solve equation $(4)$ for $u(x)$ and hence write down the solution for equation $(3)$.

I'm trying to do the first part of showing that the differential equation reduces to equation $4$. I have started out by:

\begin{align} u &= y^{-2} \\ &={1 \over y^2} \\ \therefore y^2 &= {1 \over u} \\ \implies y &= \pm {\sqrt {1 \over u}} \end{align}

I'm not sure where to continue on from here though.

Answer 1

Use the fact that

$$u=y^{-2} \implies u'=-2 y^{-3}y' \implies y'=-\frac 12 u'y^3$$

Answer 2

Substitute $$v(x)=\frac{1}{y(x)^2}$$ and $$\mu(x)=e^{\int -\frac{4}{x}ßdx}=\frac{1}{x^4}$$ then we get $$\frac{\frac{dv(x)}{dx}}{x^4}-\frac{4v(x)}{x^5}=-\frac{2}{x^6}$$ and now $$\int\frac{d}{dx}\left(\frac{v(x)}{x^4}\right)dx=\int-\frac{2}{x^6}dx$$ Can you finish?